#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <limits.h>
//宏：
//宏：定义
#define EPS 1e-9
#define MOD 100000000
//宏：类型
#define in int
#define ll long long
#define fl float
#define db double
#define ch char
#define vd void
//宏：函数
#define sc scanf
#define pr printf
//宏：其他
#define rt return
//宏：输入
#define sc_in(x) sc("%d", &(x))
#define sc_ll(x) sc("%lld", &(x))
#define sc_db(x) sc("%lf", &(x))
#define sc_ch(x) sc("%c", &(x))
#define sc_st(str) sc("%s", (str))
//宏：输出（无换行）
#define pr_in(x) pr("%d", x)
#define pr_ll(x) pr("%lld", x)
#define pr_ch(x) pr("%c", x)
#define pr_st(x) pr("%s", x)
//宏：输出（有换行）
#define pr_inn(x) pr("%d\n", x)
#define pr_lln(x) pr("%lld\n", x)
#define pr_chn(x) pr("%c\n", x)
#define pr_stn(x) pr("%s\n", x)
//宏：功能
//浮点相等判断
#define CEPS(a, b) (fabs((a) - (b)) < EPS)
//数组长度获取
#define ArrayNum(x) (sizeof(x)/sizeof(x[0]))
//函数声明：
//函数声明：输入输出辅助
//要求快速输入
in fast_read();
//要求数组输入（一维）
vd scan_arr1(in arr[], in n);
//要求数组输出（一维）
vd print_arr1(in arr[], in n);
//要求数组输入（二维）
vd scan_arr2(in arr[][100], in row, in col);
//要求数组输出（二维）
vd print_arr2(in arr[][100], in row, in col);
//函数声明：数学计算
//求某个数的阶乘（非递归）
ll factorial1(in n);
//求某个数的阶乘（递归）
ll factorial2(in n);
//求斐波那契数列第某个数（非递归）
ll fibonacci1(in n);
//求斐波那契数列第某个数（递归）
ll fibonacci2(in n);
//求斐波那契数列第某个数（递归+优化）
ll fibonacci2_memo(in n);
//求排列数P（n数，排k）
ll permute(in n, in k);
//求组合数C（n数，组k）
ll combine(in n, in k);
//求快速幂（模幂运算）
ll pow_mod(ll base, ll exp);
//求两个数的最大公因数
in gcd(in a, in b);
//求两个数的最小公倍数
in lcm(in a, in b);
//求某个数是否为素数
in is_prime(in n);
//求某个数是否为2的自然数幂
in is_power_2(in x);
//求10进制转<10进制（待拓展）
in digit_change(in n, in digit2);
//求2进制数转10进制数
in bin_to_dec(in bin);
//任意进制转换（2~36）（数组m存储结果）
in char_to_value(ch c);
ch value_to_char(in val);
in convert_base(const ch* n, in a, in b, ch m[]);
//求某个数字的反向输出
in reverse_digits(in n);
//函数声明：字符串处理
//求某个数（字符串形式）是否为回文
in is_palindrome(const ch* s);
//求某个字符串的倒置
vd reverse_string(ch* str);
//求某数是否含某数字（一般数）
in contains_digit1(in n, in p);
//求某数是否含某数字（超大数，%s输入num_str）
in contains_digit2(const ch* num_str, in p);
//求给定字符串出现频次
vd string_frequency(const ch* str);
//函数声明：数组处理
//求某十进制数的各位数
vd extract_digits(in num, in digits[]);
//函数声明：排序搜索
//要求两个整数交换
vd swap(in* a, in* b);
//要求升序排列（冒泡排序）
vd bubble_sort(in* arr, size_t n);
//要求升序排列（快速排序1到r）
vd quick_sort(in arr[], in l, in r);
//要求全排数组（start到end）
vd permute_arr(in arr[], in start, in end);
//要求翻转数组
vd reverse_arr(in arr[], in n);
//求数组某元素的位置（升序，n元，二分找x）
in binary_search(in arr[], in n, in x);
//求区间上零点的位置（单零点，二分，循环）
db binary_solve1(db low, db high);
//求区间上零点的位置（单零点，二分，递归）
db binary_solve2(db low, db high);
//求数组各元素的和
in sum_arr(in arr[], in n);
//求数组最大元素下标
in max_index(in arr[], in n);
//函数声明：几何计算
//求平面上两点的距离
db distance_2d(db x1, db y1, db x2, db y2);
//求三角形的面积（2倍）
ll triange_area_2(ll x1, ll y1, ll x2, ll y2, ll x3, ll y3);
//求凸多边形面积（2倍）
ll n_triangle_area_2(in n, ll x[], ll y[]);
//函数声明：时间日期
//求某年是否为闰年
in is_leap(in year);
//求某年某月的天数
in get_days_of_month(in year, in month);
//求某年某月某日为星期几
in get_week_day(in year, in month, in day);
//函数实现：
//函数实现：输入输出辅助
//要求快速输入
in fast_read()
{
in x = 0, f = 1;
ch c = getchar();
while (c < '0' || c > '9') {
if (c == '-') f = -f;
c = getchar();
}
while (c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = getchar();
}
return x * f;
}
//要求数组输入（一维）
vd scan_arr1(in arr[], in n)
{
for (in i = 0; i < n; i++)
{
sc("%d", &arr[i]);
}
}
//要求数组输出（一维）
vd print_arr1(in arr[], in n)
{
for (in i = 0; i < n; i++)
{
pr("%d ", arr[i], " \n"[i == n - 1]);
}
}
//要求数组输入（二维）
vd scan_arr2(in arr[][100], in row, in col)
{
for (in i = 0; i < row; i++)
{
for (in j = 0; j < col; j++)
{
while ((sc_in(arr[i][j])) != 1)
{
while (getchar() != ' ');
}
}
}
}
//要求数组输出（二维）
vd print_arr2(in arr[][100], in row, in col)
{
for (in i = 0; i < row; i++)
{
for (in j = 0; j < col; j++)
{
pr("%d ", arr[i][j]);
}
pr("\n");
}
}
//函数实现：数学计算
//求某个数的阶乘（非递归）
ll factorial1(in n)
{
if (n < 0)
return -1;
if (n > 20)
return -1;
ll result = 1;
for (in i = 2; i <= n; i++)
{
result *= i;
}
return result;
}
//求某个数的阶乘（递归）
ll factorial2(in n)
{
if (n <= 1)
{
return 1;
}
return n * factorial2(n - 1);
}
//求斐波那契数列第某个数（非递归）
ll fibonacci1(in n)
{
if (n == 0) return 0;
if (n == 1) return 1;
ll prev = 0;
ll curr = 1;
ll next;
for (in i = 2; i <= n; i++) {
next = prev + curr;
prev = curr;
curr = next;
}
return curr;
}
//求斐波那契数列第某个数（递归）
ll fibonacci2(in n)
{
if (n == 0)
{
return 0;
}
else if (n == 1 || n == 2)
{
return 1;
}
return fibonacci2(n - 1) + fibonacci2(n - 2);
}
//求斐波那契数列第某个数（递归+优化）
ll memo[1000] = { 0 };
ll fibonacci2_memo(in n)
{
if (n == 1 || n == 2)
{
return 1;
}
if (memo[n] != 0)
{
return memo[n];
}
memo[n] = fibonacci2_memo(n - 1) + fibonacci2_memo(n - 2);
return memo[n];
}
//求排列数P（n数，排k）
ll permute(in n, in k)
{
if (k < 0 || n < 0 || k > n)
{
return -1;
}
ll result = 1;
for (in i = n - k + 1; i <= n; i++)
{
result *= i;
}
return result;
}
//求组合数C（n数，组k）
ll combine(in n, in k)
{
if (k < 0 || n < 0 || k > n)
{
return -1;
}
if (k > n - k)
{
k = n - k;
}
ll result = 1;
for (in i = 0; i < k; i++)
{
result = result * (n - i) / (i + 1);
}
return result;
}
//求快速幂（模幂运算）
ll pow_mod(ll base, ll exp)
{
ll result = 1;
while (exp > 0)
{
if (exp & 1)
{
result = (result * base) % MOD;
}
base = (base * base) % MOD;
exp >>= 1;
}
return result;
}
//求两个数的最大公因数
in gcd(in a, in b)
{
if (a < 0)
a = -a;
if (b < 0)
b = -b;
while (b != 0)
{
in temp = b;
b = a % b;
a = temp;
}
return a;
}
//求两个数的最小公倍数
in lcm(in a, in b)
{
return a / gcd(a, b) * b;
}
//求某个数是否为素数
in is_prime(in n)
{
if (n <= 1)
{
return 0;
}
else if (n <= 3)
{
return 1;
}
else if (n % 2 == 0 || n % 3 == 0)
{
return 0;
}
for (in i = 5; i * i <= n; i += 6)
{
if (n % i == 0 || n % (i + 2) == 0)
{
return 0;
}
}
return 1;
}
//求某个数是否为2的自然数幂
in is_power_2(in x)
{
return x > 0 && (x & (x - 1)) == 0;
}
//求10进制转<10进制（待拓展）
in digit_change(in n, in digit2)
{
in rem = n;
in mod[100];
in i = 0;
while (rem > 0)
{
mod[i] = rem % digit2;
rem = rem / digit2;
i++;
}
in result = 0;
in ten = 1;
for (in j = 0; j < i; j++)
{
result += mod[j] * ten;
ten = 10 * ten;
}
return result;
}
//求2进制数转10进制数
in bin_to_dec(in bin) {
if (bin == 0)
{
return 0;
}
if (bin < 0)
{
return -1;
}
in decimal = 0;
in base = 1;
while (bin > 0)
{
in digit = bin % 10;
if (digit != 0 && digit != 1)
{
return -1;
}
decimal += digit * base;
base *= 2;
bin /= 10;
}
return decimal;
}
//任意进制转换（2~36）（数组m存储结果）
in char_to_value(ch c)
{
if (isdigit(c)) return c - '0';
if (isupper(c)) return c - 'A' + 10;
if (islower(c)) return c - 'a' + 10;
return -1;
}
ch value_to_char(in val)
{
if (val < 10) return '0' + val;
return 'A' + val - 10;
}
in convert_base(const ch* n, in a, in b, ch m[])
{
if (!n || !m || a < 2 || a > 36 || b < 2 || b > 36)
{
m[0] = '\0';
return -1;
}
in len = strlen(n);
if (len == 0)
{
m[0] = '0';
m[1] = '\0';
return 0;
}
ll decimal = 0;
for (in i = 0; i < len; i++)
{
in val = char_to_value(n[i]);
if (val == -1 || val >= a)
{
m[0] = '\0';
return -1;
}
decimal = decimal * a + val;
}
if (decimal == 0)
{
m[0] = '0';
m[1] = '\0';
return 0;
}
ch temp[65];
in idx = 0;
while (decimal > 0)
{
in r = decimal % b;
temp[idx++] = value_to_char(r);
decimal /= b;
}
temp[idx] = '\0';
for (in i = 0; i < idx; i++)
{
m[i] = temp[idx - 1 - i];
}
m[idx] = '\0';
return 0;
}
//求某个数字的反向输出
in reverse_digits(in n)
{
in digits[10] = { 0 };
in i = 0;
in num = n;
while (num > 0)
{
digits[i++] = num % 10;
num /= 10;
}
in ans = 0;
for (in j = 0; j < i; j++)
{
ans = ans * 10 + digits[j];
}
return ans;
}
//函数实现：字符串处理
//求某个数（字符串形式）是否为回文
in is_palindrome(const ch* s)
{
if (s == NULL)
{
return 0;
}
in left = 0;
in right = strlen(s) - 1;
while (left < right)
{
if (tolower(s[left]) != tolower(s[right]))
{
return 0;
}
left++;
right--;
}
return 1;
}
//求某个字符串的倒置
vd reverse_string(ch* str)
{
if (str == NULL)
{
return;
}
in left = 0;
in right = strlen(str) - 1;
ch temp;
while (left < right)
{
temp = str[left];
str[left] = str[right];
str[right] = temp;
left++;
right--;
}
}
//求某数是否含某数字（一般数）
in contains_digit1(in n, in p)
{
ch str[50];
ch target = '0' + p;
sprintf(str, "%d", n);
for (in i = 0; str[i] != '\0'; i++)
{
if (str[i] == target)
{
return 1;
}
}
return 0;
}
//求某数是否含某数字（超大数，%s输入num_str）
in contains_digit2(const ch* num_str, in p)
{
ch target = '0' + p;
for (in i = 0; num_str[i] != '\0'; i++)
{
if (num_str[i] < '0' || num_str[i] > '9')
{
continue;
}
if (num_str[i] == target)
{
return 1;
}
}
return 0;
}
//求给定字符串出现频次
vd string_frequency(const ch* str)
{
in freq[128] = { 0 };
for (in i = 0; str[i] != '\0'; i++)
{
unsigned ch c = str[i];
if (c == ' ' || isprint(c))
{
freq[c]++;
}
}
int count = 0;
for (in i = 0; i < 128; i++)
{
if (freq[i] > 0)
{
count++;
if (i == ' ')
{
pr("blank: %d\n", freq[i]);
}
else
{
pr("'%c': %d\n", i, freq[i]);
}
}
}
}
//函数实现：数组处理
//求某十进制数的各位数
vd extract_digits(in num, in digits[])
{
if (num == 0)
{
digits[0] = 0;
digits[1] = -1;
return;
}
in i = 0;
while (num > 0)
{
digits[i++] = num % 10;
num /= 10;
}
}
//函数实现：排序搜索
//要求两个整数交换
vd swap(in* a, in* b)
{
in temp = *a;
*a = *b;
*b = temp;
}
//要求升序排列（冒泡排序）
vd bubble_sort(in* arr, size_t n)
{
if (arr == NULL || n < 2)
{
return;
}
for (size_t i = 0; i < n - 1; i++)
{
in swapped = 0;
for (size_t j = 0; j < n - i - 1; j++)
{
if (arr[j] > arr[j + 1])
{
in temp = arr[j];
arr[j] = arr[j + 1];
arr[j + 1] = temp;
swapped = 1;
}
}
if (!swapped)
{
break;
}
}
}
//要求升序排列（快速排序1到r）
vd quick_sort(in arr[], in l, in r)
{
if (l >= r)
{
return;
}
in i = l, j = r;
in pivot = arr[(l + r) / 2];
while (i <= j)
{
while (arr[i] < pivot) i++;
while (arr[j] > pivot) j--;
if (i <= j)
{
swap(&arr[i], &arr[j]);
i++;
j--;
}
}
quick_sort(arr, l, j);
quick_sort(arr, i, r);
}
//要求全排数组（start到end）
vd permute_arr(in arr[], in start, in end)
{
if (start == end)
{
for (in i = 0; i <= end; i++)
{
pr("%d ", arr[i]);
}
pr("\n");
}
else
{
for (in i = start; i <= end; i++)
{
swap(&arr[start], &arr[i]);
permute_arr(arr, start + 1, end);
swap(&arr[start], &arr[i]);
}
}
}
//要求翻转数组
vd reverse_arr(in arr[], in n)
{
for (in i = 0; i < n / 2; i++)
{
in t = arr[i];
arr[i] = arr[n - 1 - i];
arr[n - 1 - i] = t;
}
}
//求数组某元素的位置（升序，n元，二分找x）
in binary_search(in arr[], in n, in x)
{
in left = 0, right = n - 1;
while (left <= right)
{
in mid = (left + right) / 2;
if (arr[mid] == x)
{
return mid;
}
else if (arr[mid] < x)
{
left = mid + 1;
}
else right = mid - 1;
}
return -1;
}
//求区间上零点的位置（单零点，二分，循环）
db f1(db x)
{
return sin(x);
}
db binary_solve1(db low, db high)
{
db mid;
while (high - low >= EPS)
{
mid = low + (high - low) / 2;
f1(mid) > 0 ? (high = mid) : (low = mid);
}
return low + (high - low) / 2;
}
//求区间上零点的位置（单零点，二分，递归）
db f2(db x)
{
return sin(x);
}
db binary_solve2(db low, db high)
{
db mid = low + (high - low) / 2;
if (high - low < EPS)
{
return mid;
}
else if (f2(mid) > 0)
{
return binary_solve2(low, mid);
}
else
{
return binary_solve2(mid, high);
}
}
//求数组各元素的和
in sum_arr(in arr[], in n)
{
in s = 0;
for (in i = 0; i < n; i++)
{
s += arr[i];
}
return s;
}
//求数组最大元素下标
in max_index(in arr[], in n)
{
in idx = 0;
for (in i = 1; i < n; i++)
{
if (arr[i] > arr[idx])
{
idx = i;
}
}
return idx;
}
//函数实现：几何计算
//求平面上两点的距离
db distance_2d(db x1, db y1, db x2, db y2)
{
db dx = x2 - x1;
db dy = y2 - y1;
return sqrt(dx * dx + dy * dy);
}
//求三角形的面积（2倍）
ll triange_area_2(ll x1, ll y1, ll x2, ll y2, ll x3, ll y3)
{
ll area_2 = (x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1);
return area_2 < 0 ? -area_2 : area_2;
}
//求凸多边形面积（2倍）
ll n_triangle_area_2(in n, ll x[], ll y[])
{
ll area_2 = 0;
for (in i = 0; i < n; i++)
{
in j = (i + 1) % n;
area_2 += x[i] * y[j] - x[j] * y[i];
}
return area_2 > 0 ? area_2 : -area_2;
}
//函数实现：时间日期
//求某年是否为闰年
in is_leap(in year)
{
return (year % 4 == 0) && (year % 100 != 0) || (year % 400 == 0);
}
//求某年某月的天数
in get_days_of_month(in year, in month)
{
in days = 0;
switch (month)
{
case 1: case 3: case 5: case 7: case 8: case 10: case 12:
days = 31;
break;
case 4: case 6: case 9: case 11:
days = 30;
break;
case 2:
days = is_leap(year) ? 29 : 28;
break;
}
return days;
}
//求某年某月某日为星期几
in get_week_day(in year, in month, in day)
{
in century, weekday;
if (month < 3)
{
year--;
month += 12;
}
century = year / 100;
year %= 100;
weekday = (year + year / 4 + century / 4 - 2 * century + (26 * (month + 1)) / 10 + day - 1) % 7;
if (weekday < 0)
{
weekday += 7;
}
return weekday;
}优化下界面方便我检索
本文详细介绍了如何将浮点数转换为二进制形式，包括直接使用内存中的二进制表示以及手动计算浮点数的二进制表示方法。提供了C#和C++代码示例，展示了整数部分和小数部分的转换过程。
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